Lining Up
Time Limit: 2000MS | Memory Limit: 32768K | |
Total Submissions: 19392 | Accepted: 6133 |
Description
"How am I ever going to solve this problem?" said the pilot.
Indeed, the pilot was not facing an easy task. She had to drop packages at specific points scattered in a dangerous area. Furthermore, the pilot could only fly over the area once in a straight line, and she had to fly over as many points as possible. All points
were given by means of integer coordinates in a two-dimensional space. The pilot wanted to know the largest number of points from the given set that all lie on one line. Can you write a program that calculates this number?
Your program has to be efficient!
Input
Input consist several case,First line of the each case is an integer N ( 1 < N < 700 ),then follow N pairs of integers. Each pair of integers is separated by one blank and ended by a new-line character. The input ended by N=0.
Output
output one integer for each input case ,representing the largest number of points that all lie on one line.
Sample Input
5 1 1 2 2 3 3 9 10 10 11 0
Sample Output
3
Source
给你几个点,求一条直线最多经过几个点。
个人纯暴力。
这题网上貌似说hash可以做,不懂啊,会的大神来给我讲讲啊
#include<iostream> #include<cstdio> #include<cstring> using namespace std; struct node { double x,y; } p[701]; bool v[701][701]; int main() { int n; while(cin>>n) { if(n==0)return 0; memset(v,0,sizeof(v)); for(int i=0; i<n; i++)scanf("%lf%lf",&p[i].x,&p[i].y); int a=0,b; for(int i=0; i<n-1; i++) { for(int j=i+1; j<n; j++)//我是以两个点为基点,看还有木有其他点在这两个点所构成的直线上 { if(v[i][j]==1)continue;//这是优化 b=0; //double k=(p[i].y-p[j].y)/(p[i].x-p[j].x); //double c=k*p[j].x-p[j].y; for(int u=0; u<n; u++) { if((p[u].y-p[i].y)*(p[j].x-p[i].x)==(p[j].y-p[i].y)*(p[u].x-p[i].x))//为了精度准确,尽量少用除号,我因此wa过1次 { b++; v[i][u]=1;//优化 v[u][i]=1; v[u][j]=1; v[j][u]=1;} } if(b>a)a=b; } } cout<<a<<endl; } return 0; }